water- Encoding

最新推荐文章于 2025-08-11 14:43:21 发布

weixin_34320724

最新推荐文章于 2025-08-11 14:43:21 发布

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文章标签： java

原文链接：http://www.cnblogs.com/ubuntu-kevin/archive/2013/01/28/2880649.html

本文介绍了一种字符串压缩算法，通过统计重复字符并用其出现次数加字符本身表示，从而减少原始字符串长度。每节代码示例清晰，易于理解。

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Encoding

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18521 Accepted Submission(s): 7973

Problem Description

Given a string containing only 'A' - 'Z', we could encode it using the following method:

1. Each sub-string containing k same characters should be encoded to "kX" where "X" is the only character in this sub-string.

2. If the length of the sub-string is 1, '1' should be ignored.

Input

The first line contains an integer N (1 <= N <= 100) which indicates the number of test cases. The next N lines contain N strings. Each string consists of only 'A' - 'Z' and the length is less than 10000.

Output

For each test case, output the encoded string in a line.

Sample Input

2

ABC

ABBCCC

Sample Output

ABC

A2B3C

**********************************************************************************************************

简单题，统计字母的个数。

View Code

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 char str[10005];
 5 using namespace std;
 6 int main()
 7 {
 8     int n,cnt,len;
 9     char c;
10     scanf("%d",&n);
11     getchar();
12     while(n--)
13     {
14         cnt=1;
15         fgets(str,sizeof(str),stdin);
16         c=str[0];
17         len=strlen(str);
18         for(int i=1;i<=len;i++)
19         {
20             if(c==str[i])
21                 cnt++;
22             else
23             {
24                 if(cnt==1)
25                     printf("%c",str[i-1]);
26                 else
27                     printf("%d%c",cnt,str[i-1]);
28                 c=str[i];
29                 cnt=1;
30             }
31         }
32     }
33     return 0;
34 }